An algebra problem by Hobart Pao

The basis of $\text{ker}(A)$ contains all the redundant column vectors of $A$ , namely any vector after column 2. This makes sense because any redundant column vector can be written as a non-trivial relation of the standard basis vectors. The dimension of the kernel is the number of vectors in the basis. So there are $\boxed{2016}$ of these vectors.

We can think of the matrix to be a matrix of a linear transformation( $T$ ) from a vector space $V$ of dimension $2018$ to another vector space $W$ of dimension $2017$ .

Then we know that the rank of the matrix is equal to the rank of the linear transformation.

Now we apply elementary row operations to try and make the matrix RREF ( Row Reduced Echelon Form )

apply $R_{i}=R_{i}-R_{1}$ for $i=2,3,....2017.$

and then $R_{i}=R_{i}-R_{2}$ for $i=1,3,4,....2017.$

To see that all rows after row $2$ are zero rows and it is easy to see that the rank of the matrix is 2 which is also equal to $dim(Im(T))$ where T is the linear

transformation corresponding to the given matrix.

So using Rank Nullity theorem which states that $dim(Im(T)) +dim(Ker(T))=dim(V)$ .

we see that $dim(Ker(T)) = 2018-2=2016$ .

So the kernel of the matrix also has dimension $2016$